With students beginning to attend classes across the nation, I wanted to focus the site towards some of the things they’re going to be addressing. This latest page publicize some scripts that I wrote to help with polynomial arithmetic. Originally I wrote these as homework exercises for a class in programming, but I have found them useful ever since – both in teaching mathematics classes like college algebra, which spends a lot of attention on polynomials, and in my research life. Its funny (and sad) the number of simple errors that a person (mathematician or not) can make when performing simple arithmetic, so I found it very useful to have a calculator more advanced than the simple scientific calculators that are so easily available.

I’m not going to spend a lot of time discussing the importance of polynomials, or trying to justify their need. I will bring up some problems that I’d like to address in the future, that deal with polynomials. The first is finding the roots of the characteristic polynomial of a matrix. This is useful in research because these roots are the eigenvalues of the matrix and can give many properties of the matrix. There are also some data analysis tools like Singular Value Decomposition and Principal Component Analysis where I will probably build out from this initial set of instances.

The user interface for the scripts I’ve written generate two polynomials and ask the user what is to be done with those polynomials. The options are to add the two, subtract polynomial 2 from polynomial 1, multiply the two, divide polynomial 1 by polynomial 2, and divide polynomial 2 by polynomial 1. There is also the option to make the calculations more of a tutorial by showing the steps along the way. Users who want new problems can generate a new first or second polynomial and clear work.

For addition and subtraction, the program works by first ensuring that both polynomials have the same degree. This can be achieved by adding terms with zero coefficient to the lower degree polynomial. Once this has been accomplished, we simply add the terms that have the same exponent.

For multiplication, the program first builds a matrix A, where the element ai, i+j on row i and column i+j of the matrix A is achieved by multiplying the ith term of the first polynomial by the jth term of the second polynomial. If an was not given a value in the matrix, then we put a value of zero in that cell. Once this matrix is formed, we can sum the columns of the matrix to arrive at the final answer.

The division of two polynomials works first by dividing the first term of the numerator by the first term of the denominator. This answer is then multiplied by the denominator and subtracted from the numerator. Now, the first term in the numerator should cancel and we use the result as the numerator going froward. This process is repeated as long as the numerator’s degree is still equal to or greater than the denominator’s degree.

Suppose you are at a table at a casino and notice that things don’t look quite right. Either the casino is extremely lucky, or things should have averaged out more than they have. You view this as a pattern recognition problem and would like to understand the number of ‘loaded’ dice that the casino is using and how these dice are loaded. To accomplish this you set up a number of Hidden Markov Models, where the loaded die are the latent (hidden) variables, and would like to determine which of these, if any is more likely to be using.

First lets go over a few things.

We will call each roll of the dice an observation. The observations will be stored in variables o1, o2, …, oT, where T is the number of total observations.

To generate a hidden Markov Model (HMM) we need to determine 5 parameters:

The N states of the model, defined by S = {S1, …, SN}

The M possible output symbols, defined by = {1, 2, …, M}

The State transition probability distribution A = {aij}, where aij is the probability that the state at time t+1 is Sj, given that the state at time t is Si.

The Observation symbol probability distribution B = {bj(k)} where bj(k) is the probability that the symbol k is emitted in state Sj.

The initial state distribution = {i}, where i is the probability that the model is in state Si at time t = 0.

The HMMs we’ve generated are based on two questions. For each question, you have provided 3 different answers which leads to 9 possible HMMs. Each of these models has its corresponding state transition and emission distributions.

How often does the casino change dice?

0) Dealer Repeatedly Uses Same Dice

1) Dealer Uniformly Changes Die

2) Dealer Rarely Uses Same Dice

Which sides on the loaded dice are more likely?

0) Larger Numbers Are More Likely

1) Numbers Are Randomly Likely

2) Smaller Numbers Are More Likely

How often does the casino change dice?

Which sides on the loaded dice are more likely?

(0, 0)

(0, 1)

(0, 2)

(1, 0)

(1, 1)

(1, 2)

(2, 0)

(2, 1)

(2, 2)

One of the interesting problems associated with Hidden Markov Models is called the Learning Problem, which asks the question “How can I improve a HMM so that it would be more likely to have generated the sequence O = o1, o2, …, oT?

The Baum-Welch algorithm answers this question using an Expectation-Maximization approach. It creates two auxiliary variables t(i) and t(i, j). The variable t(i) represents the probability of being in state i at time t, given the entire observation sequence. Likewise t(i, j) represents the joint probability of being in state i at time t and of being in state j at time t+1, given the entire observation sequence. They can be calculated by

Suppose you are at a table at a casino and notice that things don’t look quite right. Either the casino is extremely lucky, or things should have averaged out more than they have. You view this as a pattern recognition problem and would like to understand the number of ‘loaded’ dice that the casino is using and how these dice are loaded. To accomplish this you set up a number of Hidden Markov Models, where the loaded die are the latent variables, and would like to determine which of these, if any is more likely to be using.

First lets go over a few things.

We will call each roll of the dice an observation. The observations will be stored in variables o1, o2, …, oT, where T is the number of total observations.

To generate a hidden Markov Model (HMM) we need to determine 5 parameters:

The N states of the model, defined by S = {S1, …, SN}

The M possible output symbols, defined by = {1, 2, …, M}

The State transition probability distribution A = {aij}, where aij is the probability that the state at time t+1 is Sj, given that the state at time t is Si.

The Observation symbol probability distribution B = {bj(k)} where bj(k) is the probability that the symbol k is emitted in state Sj.

The initial state distribution = {i}, where i is the probability that the model is in state Si at time t = 0.

The HMMs we’ve generated are based on two questions. For each question, you have provided 3 different answers which leads to 9 possible HMMs. Each of these models has its corresponding state transition and emission distributions.

How often does the casino change dice?

0) Dealer Repeatedly Uses Same Dice

1) Dealer Uniformly Changes Die

2) Dealer Rarely Uses Same Dice

Which sides on the loaded dice are more likely?

0) Larger Numbers Are More Likely

1) All Numbers Are Equally Likely

2) Smaller Numbers Are More Likely

How often does the casino change dice?

Which sides on the loaded dice are more likely?

(0, 0)

(0, 1)

(0, 2)

(1, 0)

(1, 1)

(1, 2)

(2, 0)

(2, 1)

(2, 2)

One of the interesting problems associated with Hidden Markov Models is called the Decoding Problem, which asks the question “What is the most likely sequence of states that the HMM would go through to generate the sequence O = o1, o2, …, oT?

The Viterbi algorithm finds answers this question using Dynamic Programming. It creates an auxiliary variable t(i) which has the highest probability that the partial observation sequence o1, …, ot can have, given that the current state is i. This variable can be calculated by the following formula:

t(i) = maxq1, …, qt-1 p{q1, …, qt-1, qt = i, o1, …, ot | }.

1(j) = jbj(o1), for 1 j N.

Once we have calculated t(j) we also keep a pointer to the max state. We can then find the optimal path by looking at arg max 1 j NT(j) and then backtrack the sequence of states using the pointer.

Suppose you are at a table at a casino and notice that things don’t look quite right. Either the casino is extremely lucky, or things should have averaged out more than they have. You view this as a pattern recognition problem and would like to understand the number of ‘loaded’ dice that the casino is using and how these dice are loaded. To accomplish this you set up a number of Hidden Markov Models, where the number of loaded die are the latent variables, and would like to determine which of these, if any is more likely to be using.

First lets go over a few things.

We will call each roll of the dice an observation. The observations will be stored in variables o1, o2, …, oT, where T is the number of total observations.

To generate a hidden Markov Model (HMM) we need to determine 5 parameters:

The N states of the model, defined by S = {S1, …, SN}

The M possible output symbols, defined by = {1, 2, …,M}

The State transition probability distribution A = {aij}, where aij is the probability that the state at time t+1 is Sj, given that the state at time t is Si.

The Observation symbol probability distribution B = {bj(k)} where bj(k) is the probability that the symbol k is emitted in state Sj.

The initial state distribution = {i}, where i is the probability that the model is in state Si at time t = 0.

The HMMs we’ve generated are based on two questions. For each question, you have provided 3 different answers which leads to 9 possible HMMs. Each of these models has its corresponding state transition and emission distributions.

How often does the casino change dice?

0) Dealer Repeatedly Uses Same Dice

1) Dealer Uniformly Changes Die

2) Dealer Rarely Uses Same Dice

Which sides on the loaded dice are more likely?

0) Larger Numbers Are More Likely

1) All Numbers Are Randomly Likely

2) Smaller Numbers Are More Likely

How often does the casino change dice?

Which sides on the loaded dice are more likely?

(0, 0)

(0, 1)

(0, 2)

(1, 0)

(1, 1)

(1, 2)

(2, 0)

(2, 1)

(2, 2)

One of the interesting problems associated with Hidden Markov Models is called the Evaluation Problem, which asks the question “What is the probability that the given sequence of observations O = o1, o2, …, oT are generated by the HMM . In general, this calculation, p{O | }, can be calculated by simple probability. However because of the complexity of that calculation, there are more efficient methods.

The backwards algorithm is one such method (as is the forward algorithm). It creates an auxiliary variable t(i) which is the probability that the model has generated the partially observed sequence ot+1, …, oT, where 1 t T. This variable can be calculated by the following formula:

t(i) = j = 1 to N(t+1(j) * aij * bj(ot+1))

We also need that T(i) = 1, for 1 i N.

Once we have calculated the t(j) variables, we can solve the evaluation problem by p{O | } i = 1 to N1(i)

Suppose you are at a table at a casino and notice that things don’t look quite right. Either the casino is extremely lucky, or things should have averaged out more than they have. You view this as a pattern recognition problem and would like to understand the number of ‘loaded’ dice that the casino is using and how these dice are loaded. To accomplish this you set up a number of Hidden Markov Models, where the number of loaded die are the latent variables, and would like to determine which of these, if any is more likely to be using.

First lets go over a few things.

We will call each roll of the dice an observation. The observations will be stored in variables o1, o2, …, oT, where T is the number of total observations.

To generate a hidden Markov Model (HMM) we need to determine 5 parameters:

The N states of the model, defined by S = {S1, …, SN}

The M possible output symbols, defined by = {1, 2, …, M}

The State transition probability distribution A = {aij}, where aij is the probability that the state at time t+1 is Sj, given that the state at time t is Si.

The Observation symbol probability distribution B = {bj(k)} where bj(k) is the probability that the symbol k is emitted in state Sj.

The initial state distribution = {i}, where i is the probability that the model is in state Si at time t = 0.

The HMMs we’ve generated are based on two questions. For each question, you have provided 3 different answers which leads to 9 possible HMMs. Each of these models has its corresponding state transition and emission distributions.

How often does the casino change dice?

0) Dealer Repeatedly Uses Same Dice

1) Dealer Uniformly Changes Die

2) Dealer Rarely Uses Same Dice

Which sides on the loaded dice are more likely?

0) Larger Numbers Are More Likely

1) All Numbers Are Equally Likely

2) Smaller Numbers Are More Likely

How often does the casino change dice?

Which sides on the loaded dice are more likely?

(0, 0)

(0, 1)

(0, 2)

(1, 0)

(1, 1)

(1, 2)

(2, 0)

(2, 1)

(2, 2)

One of the interesting problems associated with Hidden Markov Models is called the Evaluation Problem, which asks the question “What is the probability that the given sequence of observations O = o1, o2, …, oT are generated by the HMM . In general, this calculation, p{O | }, can be calculated by simple probability. However because of the complexity of that calculation, there are more efficient methods.

The forward algorithm is one such method. It creates an auxiliary variable t(i) which is the probability that the model has generated the partially observed sequence o1, …, ot, where 1 t T. This variable can be calculated by the following formula:

t+1(j) = bj(ot+1)i = 1 to N(t(i)aij), where 1 j N, 1 t T-1.

1(j) = jbj(o1), for 1 j N.

Once we have calculated the t(j) variables, we can solve the evaluation problem by p{O | } = i = 1 to NT(i)